Known: In the representation theory of finite groups (on finite dimensional vector spaces of course), given a finite group $G$ and a representation $\rho$, we can construct an inner product $\langle \cdot,\cdot\rangle_{\rho}$ by group averaging such that our representation is unitary. In this context, unitary representations are desirable since this machinery is what is used to prove that every representation is decomposable in terms of irreducible representations.

I suspect that in the general case of a locally compact topological (Hausdorff) group $G$ and a Hilbert space $\mathcal{H}$, it is also possible but instead of averaging over the group by sums, you average over the group by an integral with respect to the Haar measure. However I get the feeling that the new inner product $\langle\cdot,\cdot\rangle_{\rho}$ would induce a different topology on $\mathcal{H}$ which is not desirable at all. Of course this doesn't happen in the finite dimensional case since all topologies would be equivalent. There might be a saving grace in this case though since all separable infinite dimensional Hilbert spaces are isomorphic (so $\mathcal{H}_{\rho}$ might be isomorphic to $\mathcal{H}$).

I'm currently reading about group C*-algebras out of Ken Davidson's text and he (and others) seem to suggest that we are particularly interested in unitary representations in $B(\mathcal{H})$ for some Hilbert space $\mathcal{H}$. In light of finite group representation theory, this is desirable.

My question then would be: how do we know we can always represent a locally compact topological (Hausdorff) group with unitaries? Clearly we want to map into $GL(\mathcal{H})$ and I suppose that by analogy to Cayley's theorem such a representation would exist. However I fail to see how we can be sure that a unitary representation exists (excluding the trivial representation since it is trivial, after all).

For what it is worth, I've noticed that representation theory courses don't always prove that a representation of a finite group exists and just assume it does. I didn't even notice that we omitted such a fact in my representation theory course until a month after final exams were over. However this sort of approach is not satisfactory to me since it is nice to know that the paradigm you're working with actually makes sense!

  • $\begingroup$ The keywords are Haar measure and Left regular representation. $\endgroup$ – Moishe Kohan Mar 14 '14 at 0:30
  • $\begingroup$ I had considered the left regular representation but I have a little bit of trouble piecing it together. Would you mind giving more details? $\endgroup$ – Cameron Williams Mar 14 '14 at 0:44
  • $\begingroup$ Maybe tomorrow. $\endgroup$ – Moishe Kohan Mar 14 '14 at 1:04

Let $G$ be a locally compact Hausdorff group and let $\mu$ be its right Haar measure. Define the complex Hilbert space $V=L^2(G, \mu)$. The group $G$ acts on $V$ via the right-regular representation $\rho$ $$ \rho_g(f)(x)= f(xg) $$ for $f\in V$ and $g, x\in G$. This representation is unitary (clear) and faithful since for every $g\in G\setminus 1$ there exists a nonempty open relatively compact subset $U\subset G$ such that $gU\cap U=\emptyset$. (Thus, $g$ does not fix the characteristic function of $U$.)

  • $\begingroup$ This is perfect! I had to prove (with some help from a friend) that you can find such a $U$ but with that fact, it's clear that the result follows. Here was my argument: Let $g_1\neq g_2\in G$ and $E\subseteq G$ be open such that $Eg_1{g_2}^{-1}\cap E=\phi$. By the claim, such an $E$ exists. Then we want to show that $\rho_{g_1}\neq\rho_{g_2}$, meaning there is an $f\in L^2(G,\mu)$ such that $\rho_{g_1}f\neq\rho_{g_2}f$. Particularly for the choice of $f = \chi_E$ this works. To see this, consider $\rho_{g_1}\chi_E - \rho_{g_2}\chi_E$. $\rho_{\cdot}$ is invertible with ... $\endgroup$ – Cameron Williams Mar 14 '14 at 17:32
  • $\begingroup$ ${\rho_g}^{-1} = \rho_{g^{-1}}$ (follows directly from definition). Then $$\rho_{g_1}\chi_E-\rho_{g_2}\chi_E = \rho_{g_1}(\chi_E-\rho_{{g_1}^{-1}}\rho_{g_2}\chi_E).$$ But $\rho_{{g_1}^{-1}}\rho_{g_2}\chi_E = \chi_{Eg_1{g_2}^{-1}}$ so we are considering $$\rho_{g_1}(\chi_E-\chi_{Eg_1{g_2}^{-1}}) \neq 0$ since the support of $E$ and $Eg_1{g_2}^{-1}$ are disjoint. $\endgroup$ – Cameron Williams Mar 14 '14 at 17:35

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